why we always choose the center of gravity of the bicycle be the rotational center.
We do not do that always, sometimes it is better to use the point in contact with the ground or some other point. We use center of mass when it leads to simpler equations than the other points. In problems dealing with torques or rotations we use the theorem:-
"The sum of torques around any stationary point $O$ equals the rate of change of the angular momentum of the body with respect to the point $O$."
In uniform gravitational field, e.g. near the ground of the Earth, the gravitational forces have zero net torque around the center of mass, so if we choose for $O$ the center of mass, the contribution to net torque due to the gravitational field vanishes and we can focus on other parts of the problem.
It is important to realize that the angular momentum with respect to point $O$ depends on where this point is. The angular momentum can be expressed as
$$
\mathbf L = \mathbf r\times M\mathbf v_{CM} + I\boldsymbol \omega
$$
where $\mathbf r$ is the radius vector of the center of mass with respect to the point $O$, $M$ is the mass of the rigid body, $\mathbf v_{CM}$ velocity of the center of mass, $I$ moment (tensor) of inertia and $\boldsymbol \omega$ angular velocity of the rotation of the body.
But if I choose the points other than the center of gravity as the rotational center, there would be net moment due to gravity
Yes.
and the rod will rotate. Will the rod rotate or not?
No. The angular momentum of the rod with respect to the point $O$ will increase, but not due to change in rotation of the rod, but just due to the fact that the velocity of the rod's center of mass $\mathbf v_{CM}$ is increasing. The rod will not change its angular velocity $\boldsymbol \omega$, since for this there would have to be non-zero net torque with respect to its center of mass (if we choose CM for $O$, the contribution $\mathbf r\times M\mathbf v_{CM}$ to angular momentum vanishes and non-zero torque then implies change in angular velocity.)
